Question: Let $x$ be a real number such that $x^3+4x=8$. Determine the value of $x^7+64x^2$.
Solution: From the equation $x^3 + 4x = 8,$ $x^3 = -4x + 8.$  Then
\begin{align*}
x^4 &= -4x^2 + 8x, \\
x^5 &= -4x^3 + 8x^2 = -4(-4x + 8) + 8x^2 = 8x^2 + 16x - 32, \\
x^6 &= 8x^3 + 16x^2 - 32x = 8(-4x + 8) + 16x^2 - 32x = 16x^2 - 64x + 64, \\
x^7 &= 16x^3 - 64x^2 + 64x = 16(-4x + 8) - 64x^2 + 64x = 128 - 64x^2.
\end{align*}Hence,
\[x^7 + 64x^2 = 128 - 64x^2 + 64x^2 = \boxed{128}.\]